Functions — f(x), composite gf(x), inverse and domain/range

Computing $gf$ when $fg$ was asked (or vice versa) — remember the function nearest the $x$ acts first.

When the variable appears twice (e.g. $y(x-2) = 3x+1$), students did not grasp the principle of factorising to collect and isolate the intended subject, so could not finish the inverse.

For an inverse found via completing the square, the answer $6 \pm \sqrt{\tfrac{x+65}{2}}$ was given without rejecting the negative root, even though the stated domain $x \ge 6$ requires only the positive one.

For $g(x)=\dfrac{5}{x-2}$, exclude $x=2$: it makes the denominator zero, so $g(2)$ is undefined. And $x=2$ can never be a valid solution of $g(x)=k$ — but $x=0$ here is fine.

Completing the square to invert $2x^2-24x+7$ gives $2(x-6)^2-65$, since $2\times 6^2 = 72$. Writing $-36$ instead of $-72$ (forgetting to multiply the $6^2$ by the $2$) is a common slip.